Testing hypothesis with two proportions and Chi-square testing

Hypothesis testing for homogeneity of proportions We wish to compare the

effects of two different insecticides. The first insectide, InsectsRUs, was applied to 300 insects and 50 of them died. The second insecticide, InsectsBGone, was applied to 350 insects and 90 died. Test to see if there is a significant difference in the proportion of insects killed.

Hypothesis of interest

H0: p1 = p2 versus HA: p1p2

Test statistic:

Where

1 2

1 2

ˆ ˆ

1 1ˆ ˆ(1 )C C

p p

p pn n

1 2

1 2

ˆCX X

pn n

Calculations

22

2

90ˆ 0.25714

350

Xp

n

InsectsRUs

InsectsBGone

p-value=0.00515

1 2

1 2

ˆ ˆ

1 1ˆ ˆ(1 )C C

p p

p pn n

1 2

1 2

ˆCX X

pn n

11

1

50ˆ 0.166667

300

Xp

n

50 90 140ˆ 0.21538

350 300 650Cp

0.1667 0.25712.79719

1 10.2153(1 0.2153)

300 350

Contingency Table Can also view this as a contingency table

Type of InsecticideInsectsRUs InsectsBGone

Killed 50 90Not killed 250 260

TOTAL 300 350

Chi-square Chi-square testing can

be used to test if the distribution is the same in each group (i.e. insecticides)

Need to find Expected values

Insecticide

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Killed Not Killed

InsectRUs

InsectBGone

Expected Values for Chi Square Expected value =

(row total)*(column total)/N

InsectsRUs

InsectsBGone TOTAL

Killed 50 90 140

Not Killed 250 260 510

TOTAL 300 350 650Column Totals

Row Totals

N

InsectsRUs InsectsBGone TOTAL

Killed 50 90 140

Not Killed 250 260 510

TOTAL 300 350 650

InsectsRUs InsectsBGone TOTAL

Killed =140*300/650 =140*350/650 140

Not Killed =510*300/650 =350*510/650 510

TOTAL 300 350 650

Calculating Expected Values

Calculated expected values

InsectsRUs InsectsBGone TOTAL

Killed 64.61538 75.38462 140

Not Killed 235.38462 274.61538 510

TOTAL 300 350 650

22 (observed-expected)

expectedX Now calculate the Chi –square statistic

2 2 2 22 (50 64.61538) (90 75.38462) (250 235.38462) (260 274.61538)

64.61538 75.38462 235.38462 274.61538X

=7.824768 , degrees of freedom = 1

p-value = 0.00515

Comments on the Chi-square Test The test of H0: no association versus HA: there is an

association between two categorical variables is computationally the same as testing if the conditional distributions are the same (this can be extended to more than two populations).

The degrees of freedom for a chi-square test is (r-1)*(c-1), where r = # rows and c=# columns.

Be cautious when expected frequencies are lower than 5.

The chi-square test can also be used to test for goodness-of-fit.

Goodness of fit Test to see if percent of longleaf pine is evenly distributed

across the four quadrants (example from Moore, McCabe, and Craig).

Quad1 Quad2 Quad3 Quad4

Count 18 22 39 21

There are a total of 100 trees, so we would expect ¼ of 100 to be in each quadrant (where expected value =0.25*100=25).

2 2 2 2 22 (observed-expected) (18 25) (22 25) (39 25) (21 25)

expected 25 25 25 25X

=10.8, degrees of freedom=3

P-value=0.012858